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Let $S = \{4 - t, t^3, 6t^2, 3t + t^3, -1 + 4t\}$. Is $S$ a basis for $P^3$?

Polynomial Equation:
0t3 + 0t2 - t + 4 = 0
t3 + 0t2 + 0t + 0 = 0
0t3 + 6t2 + 0t + 0 = 0
t3 + 0t2 + 3t + 0 = 0
0t3 + 0t2 + 4t - 1 = 0

Transform into an augmented matrix:
\begin{bmatrix}0&0&-1&4&0\\1&0&0&0&0\\0&6&0&0&0\\1&0&3&0&0\\0&0&4&-1&0\end{bmatrix} RREF:
\begin{bmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0\end{bmatrix}

I answered the question saying that S is not a basis for p3 because the all zero bottom row means that there is a free variable within the equation, making the overall system linearly dependent and therefore not a basis.

To push my understanding, can I further say that the vector (-1 + 4t) is the only dependent variable and if it was not in S, then S would be a basis of p3?

Also, the presence of the free variable would mean that the dependent vector would be a non-unique combination of ANY of the other vectors found within S?

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$P^3$ is 4 dimensional with a natural basis, $1, t, t^2, t^3$. In this basis, we can represent $S$ as the set of columns of

\begin{pmatrix} 4 & 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 3 & 4 \\ 0 & 0 & 6 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ \end{pmatrix}

This row reduces to

\begin{pmatrix} 1 & 0 & 0 & 0 & -\frac{1}{4} \\ 0 & 1 & 0 & 0 & -\frac{5}{4} \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & \frac{5}{4} \\ \end{pmatrix}

Row reducing gives us a basis for the relations between columns (the null space). Specifically, we have one relation

$$ (4 - t) + 5 (t^3) - 5(3t + t^3) + 4(-1 + 4t) = 0 $$

I can solve this equation for any element of $S$ except $6t^2$. This means that when I remove $4-t$ or $t^3$ or $3t+t^3$ or $-1 + 4t$ from $S$, I get a basis (because I no longer have the above relation and because I have a way of writing the removed vector in terms of the remaining vectors [using the above relation]).

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  • $\begingroup$ Okay, great. Thanks for the clarification $\endgroup$ – Evan Kim Nov 6 '18 at 18:27
  • $\begingroup$ @Evan If an answer is useful to you (and solves your problem), you should upvote it and accept it. This is optional of course, but it helps future visitors to identify the most helpful answers. $\endgroup$ – Trevor Gunn Nov 6 '18 at 19:55
  • $\begingroup$ okay i did it, thanks again $\endgroup$ – Evan Kim Nov 6 '18 at 20:02
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Good for you trying to deepen your understanding of this subject. Unlike calculus, where you're usually just studying one thing at a time (functions), linear algebra has a lot of objects with different properties. These objects are related, and sometimes in a canonical fashion, but they're not the same object. Sets of vectors are not identical to matrices, matrices are not identical to systems of equations, variables are not the same as vectors, etc. So I want to pick apart some of your statements to make sure you're using the terms correctly.

I set all the polynomials = 0, and reduced the augmented 5x5 matrix down to reduced row echelon form

To push your understanding: why did you do this? If your answer is “Aren't you supposed to do that?” then you should dig a little deeper. You didn't show your augmented matrix, but based on its dimensions I'm assuming you put the coefficients of the polynomials as rows of the matrix. But the more direct connection to linear dependence is the matrix whose columns are the coefficients of the polynomials. If $A$ is that matrix, then a linear dependence relation among the vectors in $S$ is a solution to the linear system $Ax =0$. Note that $A$ has dimensions $4\times 5$. If you form the augmented matrix by adding the column of zeros, that augmented matrix has dimensions $4\times6$.

the all zero bottom row means that there is a free variable within the equation,

Ask yourself: What equation? Usually a row of zeroes in the RREF means that you have some redundancy in the system of equations, not necessarily a free variable. For instance, the system $x=1$ and $3x=3$ will have a row of zeroes in the RREF of the augmented matrix, but there aren't any free variables.

Again, using the vectors in $S$ as columns of $A$ gives you the answer you want: The RREF of $A$ will have a column of zeroes. So there is a free variable in the system $Ax=0$, that is, there are nontrivial solutions to the system.

making the overall system linearly dependent and therefore not a basis.

A system of equations cannot be linearly dependent, only a set of vectors can be linearly dependent. Likewise, a system of equations cannot be a basis, only a set a vectors can be a basis. In this case, the nontrivial solutions to $Ax=0$ do imply that $S$ (not “the system”) is linearly dependent, and therefore not a basis.

can I further say that the vector $(-1 + 4t)$ is the only dependent variable

A vector is not a variable, dependent or otherwise. In this system $Ax=0$, the last column of $A$ corresponds to the vector $-1+4t$, and the last variable $x_5$ is free. You're eliding this correspondence when you use the word “is”. But I think it's important to understand that vectors and variables are fundamentally different objects.

But $-1+4t$ is not unique even in this regard. If you reversed the order of the vectors, you'd still get a free variable in the associated system, but this time it would correspond to the vector $4-t$.

and if it was not in S, then S would be a basis of $P^3$?

Now that is a good question! If you let $S'$ be the set with $-1+4t$ removed, and $A'$ the matrix whose columns are the vectors in $S'$, you'll find that the RREF of $A'$ is $I_4$. So yes, $S'$ is a basis of $P^3$. Does every vector in $S$ have that property, that if you remove it from $S$, what's left is a basis?

Also, the presence of the free variable would mean that the dependent vector would be a non-unique combination of ANY of the other vectors found within S?

Sort of. If $Ax = 0$, then each of the columns in $A$ corresponding to nonzero components of $x$ can be written as a linear combination of the other columns (again, corresponding to nonzero components). In the example above, you'll never get the column of $A$ corresponding to the vector $6t^2$ to be zero in the RREF. So you'll never be able to write that vector as a linear combination of the others.

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  • $\begingroup$ Hey thanks for the lengthy explanation, this is what I was looking for as I feel I cannot properly speak the lingo yet! Please see my original post, I edited in how I set the system up and what the augmented matrix looks like. Matrix A is actually a 5x4 and the augmented matrix makes it a 5x5 matrix. I found your comments about the rows vs columns interesting, however I wasn't able to follow your thought process. Could you walk through it again? On a side note, I thought that they are the same since that a column = row transposed and vice versa. $\endgroup$ – Evan Kim Nov 7 '18 at 0:38
  • $\begingroup$ @EvanKim: thanks, I'm glad what I've written so far is useful. What you edited in is how I surmised you were thinking. Hopefully I can come back later and expand my answer to address it. But briefly: in your "system" where you set all the polynomials equal to zero, where are the unknowns? You only have one variable in the five equations ($t$), and that isn't so much an unknown as a differentiator between the coefficients. $\endgroup$ – Matthew Leingang Nov 7 '18 at 12:43
  • $\begingroup$ I think I see now the biggest flaw in my reasoning thus far was thinking that a variable like t squared was different from t. I will have to double check my textbook when I get home for the exact wording as I made a hasty assumption that t1, t2, t3 were different variables, not the same variable with different exponents! Once I do that, I will re-read through your original post and then comment back again with my findings. $\endgroup$ – Evan Kim Nov 8 '18 at 14:16
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A basis for a vector space of dimension n must satisfies three properties: 1) The vectors span the space. 2) The vectors are independent. 3) There are n vectors in the space. And any two are sufficient to show the third!

$P^3$, the space of all third degree polynomials, has dimension 4 (the "standard basis is $\{1, t, t^2, t^3\}$. The given set has five vectors so cannot possibly be a basis!

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  • $\begingroup$ Yes, one of the five vectors is linearly dependent to one of the four other vectors. If that linearly dependent vector, which is is (-1 + 4t) is not part of the system, then S will be a basis? That is what I want to confirm $\endgroup$ – Evan Kim Nov 6 '18 at 18:25

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